Differential equation with the Substitution method I'm trying to solve this equation below :
$$ y' = \sin\left(\frac{y}{x}\right) + \frac{y}{x}  \tag 1 $$
The first step was to to substitute $\frac{y}{x}$ with $u$
=>   $ u = \frac{y}{x}$     => $y' = u'x + u \tag 2$
Then with
$(1) = (2)$,  I got at the end
$$\dfrac{\ln\left(\cos\left(y\right)+1\right)-\ln\left(1-\cos\left(y\right)\right)}{2} = \ln|x| + C$$
And after more simplification :
$$\frac{\cos(y)}{1-\cos(y)} = 2Cx   $$
How can I get the general function from this equation ?
 A: From your last line:
$$\frac{\cos(y)+1}{1-\cos(y)} = Cx^2$$
$$\cos(y)+1 = Cx^2(1- \cos y)$$
$$ \cos y (1+Cx^2)=Cx^2-1$$
$$ \cos y =\dfrac {Cx^2-1}{ Cx^2+1}$$
$$\implies y(x)=\cos^{-1} \left ( \dfrac {Cx^2-1}{ Cx^2+1} \right)$$

Or more simply:
$$u'x= \sin u$$
$$\int \dfrac {du}{\sin u}=\int \dfrac {dx}{x}$$
$$-\ln (\cot u/2)=\ln x +K$$
$$\cot \dfrac u 2=\dfrac C x$$
$$u=2 \cot ^{-1} \left ( \dfrac Cx \right)$$
$$\boxed {y(x)=2x \cot ^{-1} \left ( \dfrac Cx \right)}$$

For the integral note that:
$$I=\int \dfrac {dx}{\sin x }=\int \dfrac {dx}{2 \sin x/2 \cos x/2}$$
$$I=\int \dfrac 12\dfrac {\cos x/ 2dx}{\sin x/2 \cos ^2 x/2}$$
$$I=\int \dfrac {d \tan \dfrac x 2}{\tan \dfrac x2}=\ln \tan \dfrac x 2+C$$
Or you can use Weierstrass substitution  
$$\int \dfrac {dx}{\sin x} =\int \dfrac {1+t^2}{2t}\dfrac {2dt}{1+t^2}
=\int \dfrac {dt}{t}=\ln t +C$$
Where $t=\tan \dfrac x 2$
A: There is an amazing manner for solving
$$y' = \sin\left(\frac{y}{x}\right) + \frac{y}{x}  $$
First use $y=x\,z$ to make
$$x\,z'=\sin(z)$$ Now, switch variables
$$\frac x {x'}=\sin(z) \implies \frac{x'}x=\csc(z)\implies \log(|x|)+c=\log \left(\tan \left(\frac{z}{2}\right)\right)$$
